 Hi and welcome to the session. Let's work out the following question. The question says if the radii of the ends of a bucket 45 centimeter high are 28 centimeter and 7 centimeter Determine the capacity and total surface area of the bucket So let's start with the solution to this question We see that this bucket is actually a frustrum So let the small radius of the frustrum that is R2B7 centimeter and the larger radius that is R1B28 centimeter Height of the bucket is given to be 45 centimeter Therefore Volume of the bucket That is V is equal to 1 by 3 into pi into H into R1 square plus R2 square plus R1 R2 We simply put in the values here and we get 1 by 3 into 22 by 7 into 45 into 7 square plus 28 square plus 7 into 28 That is equal to 22 by 7 into 15 into 1029 And this is equal to 48510 centimeter cube Now slant high that is L of this frustrum is given by square root of R1 minus R2 the whole square plus H square This is equal to square root of 28 minus 7 the whole square plus 45 square This is equal to square root of 21 square plus 45 square Which is equal to square root of 441 plus 2025 That is equal to 49.66 Therefore Total surface area of this bucket will be equal to curved surface area plus Area base surface area is given by pi into R1 plus R2 into L That is equal to 22 by 7 into 28 plus 7 into 49.66 That is equal to 22 by 7 into 35 into 49.66 That is equal to 5462.6 centimeter square Also Area of the base is equal to pi that is 22 by 7 into 7 into 7 That is equal to 22 into 7 Which is equal to 154 centimeter square Therefore The total surface area is equal to 5462.6 plus 154 centimeter square That is equal to 5616.6 centimeter square So our answer to this question is The capacity of the bucket is 48510 centimeter cube And the surface area of the bucket is 5616.6 centimeter square So this is our answer to this question I hope that you understood the solution I enjoyed the session Have a good day